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What is the smallest integer you can create using 1,2,3,4,5,6,7,8,9 that is strictly greater than 1?

  • You can use any mathematics symbol.
  • You must use all nine numbers.
  • You can't use the same number more than once.
  • You can't use the same mathematics symbol more than once.
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closed as too broad by Bass, ffao, JonMark Perry, A. P., Rubio Mar 8 '18 at 20:15

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ "You can use any mathematics symbol" - you need to narrow this to a finite list. Otherwise I can just invent a mathematical symbol $*$ and define it almost however I want. There are a LOT of mathematical symbols out there. $\endgroup$ – Rand al'Thor Mar 8 '18 at 11:10
  • $\begingroup$ There's nothing wrong with the distinction "counting number" which refers to the natural numbers, i.e. positive integers. Here they are equivalent, since we only consider numbers greater than 1. I don't think there was any ambiguity in that part of the original statement. $\endgroup$ – AlexanderJ93 Mar 8 '18 at 11:43
  • $\begingroup$ Adding to @votbear 's answer, if the smallest number we could make it not strictly greater than $1$, but still is greater nonetheless, then it would be as follows: $$\frac{1}{2\times 3\times 4\times 5\times 6\times 7\times 8\times 9} = \frac{1}{9!}$$ Through multiplication, it will be using $\times$ more than once, but we can write the denominator as $9!$ which is an alternative mathematical symbol that is only used once so.... $\endgroup$ – Mr Pie Mar 19 '18 at 10:10
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Uh...

The smallest integer which is strictly greater than 1 is 2, right?

Which is achievable by many means, one of which is

2 + (18 / 3 - 6) * [the rest of the numbers here]

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  • $\begingroup$ You can't use the same mathematics symbol more than once. How would you add the rest of the numbers ? $\endgroup$ – Doomenik Mar 8 '18 at 12:33
  • $\begingroup$ @Doomenik By concatenating the remaining numbers to 14579. That is if concatenating is allowed, otherwise this solution is invalid. $\endgroup$ – Rick van Osta Mar 8 '18 at 12:53

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